On an exercise from Hartshorne's Algebraic Geometry

My question is in fact the exercise 1.8 page 8 in the book GTM52 by Robin Hartshone.

Let $Y$ be an affine variety of dimension $r$ in $\mathbb{A}^n$. Let $H$ be a hypersurface in $\mathbb{A}^n$ and assume that $Y\nsubseteq H$. Then prove that every irrducible component of $Y\cap H$ has dimension $r-1$

Here is my idea to prove :

Suppose that $Y\cap H =\cup U_i$ then $I(Y\cap H)=I(\cup U_{i})$. It is not hard to check that $f \in I(Y\cap H)$ then $(f)\subseteq I(U_{i})$ for each $i$. Since $U_i$ is irreducible then $I(U_{i})$ is a prime ideal.

Then if $I(U_i)$ is minimal over $(f)$ then we can use Krull Principal Ideal Theorem to prove that ht$I(U_i)=1$, then dim$U_i=r-1$

I have two question :

• Is my strategy wrong ?
• If I am on the right way, how can I show that $I(U_{i})$ is minimal over $(f)$ ?

Thank for reading my question.

Let $f$ be a regular function defining the hypersurface $H$ and suppose the $U_i$ are the irreducible components of $Y\cap H$ (otherwise the result is not true). If $I(U_i)$ is not minimal over $(f)$, it contains strictly a prime ideal $J\ni f$. So $U_i\subsetneq Z(J) \subseteq Z(f)=H$. As $Z(J)$ is irreducible, this contradicts the hypothesis that $U_i$ is an irreducible component of $H\cap Y$.